问题描述:
Implement int sqrt(int x).
Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842…, and since the decimal part is truncated, 2 is returned.
题源:here;完整实现:here
思路:
三种方案:1 直接调用库函数;2 牛顿方法;3 二分法。从时间上,三种方式很接近。
1 调用库函数
int mySqrt(int x) {
return int(sqrt(x));
}
运行结果:
2 牛顿方法
int mySqrt2(int x){
if (x <= 0) return 0;
double last = 0, rest = 1;
while (last != rest){
last = rest; rest = (rest + x / rest) / 2;
}
return int(last);
}
运行结果:
3 二分法
int mySqrt3(int x){
if (x <= 0) return 0;
int left = 0, right = x/2 + 1;
while (left < right){
long long mid = (left + right + 1) / 2;
long long sq = mid*mid;
if (sq == x) return int(mid);
else if (sq < x) left = mid;
else right = mid - 1;
}
return left;
}
运行结果: